Question: Simplify and expand the following expression: $ \dfrac{r}{r - 3}-\dfrac{9}{2r + 4} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(r - 3)(2r + 4)$ Multiply the first term by $\dfrac{2r + 4}{2r + 4}$ $ \begin{align*} \dfrac{r}{r - 3} \times \dfrac{2r + 4}{2r + 4} & = \dfrac{(r)(2r + 4)}{(r - 3)(2r + 4)} \\ & = \dfrac{2r^2 + 4r}{(r - 3)(2r + 4)}\end{align*} $ Multiply the second term by $\dfrac{r - 3}{r - 3}$ $ \begin{align*} \dfrac{9}{2r + 4} \times \dfrac{r - 3}{r - 3} & = \dfrac{(9)(r - 3)}{(2r + 4)(r - 3)} \\ & = \dfrac{9r - 27}{(2r + 4)(r - 3)}\end{align*} $ Now we have: $ = \dfrac{2r^2 + 4r}{(r - 3)(2r + 4)} - \dfrac{9r - 27}{(2r + 4)(r - 3)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2r^2 + 4r - (9r - 27)}{(r - 3)(2r + 4)} $ $ = \dfrac{2r^2 + 4r - 9r + 27}{(r - 3)(2r + 4)} $ $ = \dfrac{2r^2 - 5r + 27}{(r - 3)(2r + 4)}$ Expand the denominator: $ = \dfrac{2r^2 - 5r + 27}{2r^2 - 2r - 12}$